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Section 7.1 Integration by Parts

Suppose u and v are functions of \(x\text{.}\)

Recall the produce rule:

\begin{equation*} \dfrac{d}{dx}(uv) = u \dfrac{d}{dx} + u \dfrac{d}{dx} \end{equation*}

We can take the integtal with respect to \(x\) of both sides of the above formula to get:

\begin{align*} \int \dfrac{d}{dx}(uv)dx \amp = \int u \dfrac{d}{dx} + u \dfrac{d}{dx} dx \\ uv \amp = \int u \dfrac{dv}{dx}dx + \int v \dfrac{du}{dx} dx \\ uv \amp = \int udv + \int v du \end{align*}

Note 7.1.1.
Notice that the dx's cancel.

If re rearange the terms we get the integration by parts formula:

\begin{equation*} \int u dv = uv - \int v du \end{equation*}
Example 7.1.1.

SSolve using integration by parts: \(\int x \cos(x) dx\)

Answer

\begin{align*} \int x \cos(x) dx \amp = x \sin(x) - \int \sin(x) \cdot 1 dx\\ \amp = x \sin(x) - ( - \cos(x)) + c \\ \amp = x \sin(x) + \cos(x) + c \end{align*}
\begin{equation*} \begin{cases} \text{}\\ u = x \\ v = \sin(x) \\ du = 1 dx \\ dv = \cos(x) dx \end{cases} \end{equation*}

Rule of Thumb.

From left to right:

logrithmic, inverse trig, algebraic, trig, exponential

Where,

  • u is the earliest in the list

  • dv is the latest in the list

This doesn't awlays work, but works enough to be helpful.

Subsection 7.1.1 Tabular Integration

A quick way to integrate where you would otherwise have to integrate by parts multiple times. You do this by differenciating the u until it reaches zero and then taking the integral of the dv the same number of times.

Table 7.1.2. Tabular Integration
derivative integral
+ \(x^3 + 5x^2 + 7x + 4\) \(e^{2x}\)
- \(3x^2 + 10x + 7\) \(\dfrac{1}{2} e^{2x}\)
+ \(6x + 10\) \(\dfrac{1}{4} e^{2x}\)
- \(6 \) \(\dfrac{1}{8}e^{2x}\)
\(0 \) \(\dfrac{1}{16}e^{2x}\)
The sign in front of each line denotes weather it is positive or negative. We multiply the polynomial on the derivative column by the integral on the next row. You do this until you reach the zero. After that just add a "\(+ c\)" to the end.

\begin{equation*} \int (x^3 + 5x^2 + 7x + 4) \cdot 3^{2x} dx = (x^3 + 5x^2 + 7x + 4) \cdot \dfrac{1}{2} e^{2x} - (3x^2 + 10x + 7) \cdot \dfrac{1}{4} e^{2x} + (6x+10) \cdot \dfrac{1}{8} e^{2x} - \dfrac{6}{16}e^{2x} + c \end{equation*}

What would we do if u is not a polynomial or if repeated differentiation of u doesn't eventually give 0? For example \(\int \ln(x)dx\text{:}\)

Table 7.1.3. Tabular Integration (Repeating)
derivative integral
+ \(\ln(x)\) \(1\)
- \(\dfrac{1}{x}\) \(x\)
\(\vdots\) \(\vdots\)
When we have this type of integral, we can stop when we know how to integrate one element in the row by the other element. In this case, the second row is \(\dfrac{1}{x}\) time \(x\text{,}\) which is just 1.

\begin{align*} \int \ln(x) dx \amp = \ln(x) \cdot x - \int \dfrac{1}{x} x dx \\ \amp = x \ln(x) - x + c \end{align*}

Subsection 7.1.2 A Special Integration Technique

There is a special integration technique used where the original integral appears in the answer when integrating by parts.

\begin{align*} \int \sin (x) \cdot e^x dx \amp = \sin(x) \cdot e^x - \int \cos (x) \cdot e^x dx \\ \amp = \sin(x) \cdot e^x - ( \cos(x) \cdot e^x - \int e^x ( - \sin(x) )dx)\\ \amp = \sin(x) \cdot e^x - \cos(x) \cdot e^x - \int \sin(x) \cdot e^x dx \end{align*}

Table 7.1.4.
\(u = \sin(x)\) \(v = e^x\)
\(du = \cos(x)dx\) \(dv = e^xdx\)
Table 7.1.5.
\(u = \cos(x)\) \(v = e^x\)
\(du = - \sin(x)dx\) \(dv = e^xdx\)

But, if we add \(\int \sin(x) \cdot e^x dx\) to both sides, we can get a reasonable answer:

\begin{align*} \int \sin (x) \cdot e^x dx \amp = \sin(x) \cdot e^x - \cos(x) \cdot e^x - \int \sin(x) \cdot e^x dx \\ 2 \int \sin (x) \cdot e^x dx \amp = \sin(x) \cdot e^x - \cos(x) \cdot e^x \\ \int \sin (x) \cdot e^x dx \amp = \dfrac{1}{2} ( \sin(x) \cdot e^x - \cos(x) \cdot e^x) + c \end{align*}